The $sigma$-algebra generated by $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$...












2














I am trying to show that the $sigma$-algebra generated by $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$ contains all singleton sets ${x} in mathbb{R}$.



I know that $F$ is an algebra. I believe $F$ itself is not a $sigma$-algebra, since, if we take an infinite sequence of $A_n in F$ such that $0 in ((A_n)^c)^circ$ for each $n in mathbb{N}$ then we do not necessarily have that $0 in cup_{n=1}^{infty}A_n$ (since $0 in cap_{n=1}^{infty}(A_n^c)^circ supseteq (cap_{n=1}^{infty}((A_n)^c)^circ = (cup_{n=1}^{infty}A_n)^circ$).



So, I am not exactly sure how to explicitly construct and/or describe the $sigma$-algebra generated by $F$. (From there, I am pretty sure I should be able to show that the singleton sets are in $sigma(F)$. I just don't know how to get to that point in the first place.)



This link gave an explanation about constructing a $sigma$-algebra from a collection of sets, but it ended up being more confusing than helpful.






real-analysis elementary-set-theory







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  • $F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
    – DanielWainfleet
    Nov 25 at 20:08












  • @DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
    – Jane Doe
    Nov 25 at 20:11


















2














I am trying to show that the $sigma$-algebra generated by $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$ contains all singleton sets ${x} in mathbb{R}$.



I know that $F$ is an algebra. I believe $F$ itself is not a $sigma$-algebra, since, if we take an infinite sequence of $A_n in F$ such that $0 in ((A_n)^c)^circ$ for each $n in mathbb{N}$ then we do not necessarily have that $0 in cup_{n=1}^{infty}A_n$ (since $0 in cap_{n=1}^{infty}(A_n^c)^circ supseteq (cap_{n=1}^{infty}((A_n)^c)^circ = (cup_{n=1}^{infty}A_n)^circ$).



So, I am not exactly sure how to explicitly construct and/or describe the $sigma$-algebra generated by $F$. (From there, I am pretty sure I should be able to show that the singleton sets are in $sigma(F)$. I just don't know how to get to that point in the first place.)



This link gave an explanation about constructing a $sigma$-algebra from a collection of sets, but it ended up being more confusing than helpful.






real-analysis elementary-set-theory







share|cite|improve this question
























  • $F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
    – DanielWainfleet
    Nov 25 at 20:08












  • @DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
    – Jane Doe
    Nov 25 at 20:11
















2












2








2







I am trying to show that the $sigma$-algebra generated by $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$ contains all singleton sets ${x} in mathbb{R}$.



I know that $F$ is an algebra. I believe $F$ itself is not a $sigma$-algebra, since, if we take an infinite sequence of $A_n in F$ such that $0 in ((A_n)^c)^circ$ for each $n in mathbb{N}$ then we do not necessarily have that $0 in cup_{n=1}^{infty}A_n$ (since $0 in cap_{n=1}^{infty}(A_n^c)^circ supseteq (cap_{n=1}^{infty}((A_n)^c)^circ = (cup_{n=1}^{infty}A_n)^circ$).



So, I am not exactly sure how to explicitly construct and/or describe the $sigma$-algebra generated by $F$. (From there, I am pretty sure I should be able to show that the singleton sets are in $sigma(F)$. I just don't know how to get to that point in the first place.)



This link gave an explanation about constructing a $sigma$-algebra from a collection of sets, but it ended up being more confusing than helpful.






real-analysis elementary-set-theory







share|cite|improve this question















I am trying to show that the $sigma$-algebra generated by $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$ contains all singleton sets ${x} in mathbb{R}$.



I know that $F$ is an algebra. I believe $F$ itself is not a $sigma$-algebra, since, if we take an infinite sequence of $A_n in F$ such that $0 in ((A_n)^c)^circ$ for each $n in mathbb{N}$ then we do not necessarily have that $0 in cup_{n=1}^{infty}A_n$ (since $0 in cap_{n=1}^{infty}(A_n^c)^circ supseteq (cap_{n=1}^{infty}((A_n)^c)^circ = (cup_{n=1}^{infty}A_n)^circ$).



So, I am not exactly sure how to explicitly construct and/or describe the $sigma$-algebra generated by $F$. (From there, I am pretty sure I should be able to show that the singleton sets are in $sigma(F)$. I just don't know how to get to that point in the first place.)



This link gave an explanation about constructing a $sigma$-algebra from a collection of sets, but it ended up being more confusing than helpful.






real-analysis elementary-set-theory




real-analysis elementary-set-theory






share|cite|improve this question















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edited Nov 25 at 20:09

























asked Nov 25 at 19:13









Jane Doe

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  • $F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
    – DanielWainfleet
    Nov 25 at 20:08












  • @DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
    – Jane Doe
    Nov 25 at 20:11




















  • $F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
    – DanielWainfleet
    Nov 25 at 20:08












  • @DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
    – Jane Doe
    Nov 25 at 20:11


















$F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
– DanielWainfleet
Nov 25 at 20:08






$F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
– DanielWainfleet
Nov 25 at 20:08














@DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
– Jane Doe
Nov 25 at 20:11






@DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
– Jane Doe
Nov 25 at 20:11












2 Answers
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It contains ${0}$ because that is the intersection of the sets $A_n = (-frac{1}{n},frac{1}{n})$, which all verify $0in A_n^circ$, hence $A_n in F$. On the other hand, if $xinmathbb{R}setminus {0}$, then we have that ${x}$ is closed and ${x}^c$ is an open set containing $0$, hence ${x} in F$.






share|cite|improve this answer



















  • 2




    This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
    – Andreas Blass
    Nov 26 at 1:35












  • You are right, I have edited my solution to make it simpler according to your remark.
    – mlerma54
    Nov 26 at 3:30



















1














Every $Ssubset Bbb R$ belongs to $sigma(F).$



(I)... $Ssetminus (-1/n,1/n)in F$ for each $nin Bbb N$ so $Ssetminus {0}=cup_{nin Bbb N},(,Ssetminus (-1/n,1/n),)in sigma (F).$



(II)... $(-1/n,1/n)in F$ for each $nin Bbb N$ so ${0}=cap_{nin Bbb N}(-1/n,1/n)in sigma (F).$



(III)... So either $S=Ssetminus {0}in sigma(F)$ or $S=(Ssetminus {0})cup {0}in sigma(F).$






share|cite|improve this answer





















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    2 Answers
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    It contains ${0}$ because that is the intersection of the sets $A_n = (-frac{1}{n},frac{1}{n})$, which all verify $0in A_n^circ$, hence $A_n in F$. On the other hand, if $xinmathbb{R}setminus {0}$, then we have that ${x}$ is closed and ${x}^c$ is an open set containing $0$, hence ${x} in F$.






    share|cite|improve this answer



















    • 2




      This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
      – Andreas Blass
      Nov 26 at 1:35












    • You are right, I have edited my solution to make it simpler according to your remark.
      – mlerma54
      Nov 26 at 3:30
















    2














    It contains ${0}$ because that is the intersection of the sets $A_n = (-frac{1}{n},frac{1}{n})$, which all verify $0in A_n^circ$, hence $A_n in F$. On the other hand, if $xinmathbb{R}setminus {0}$, then we have that ${x}$ is closed and ${x}^c$ is an open set containing $0$, hence ${x} in F$.






    share|cite|improve this answer



















    • 2




      This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
      – Andreas Blass
      Nov 26 at 1:35












    • You are right, I have edited my solution to make it simpler according to your remark.
      – mlerma54
      Nov 26 at 3:30














    2












    2








    2






    It contains ${0}$ because that is the intersection of the sets $A_n = (-frac{1}{n},frac{1}{n})$, which all verify $0in A_n^circ$, hence $A_n in F$. On the other hand, if $xinmathbb{R}setminus {0}$, then we have that ${x}$ is closed and ${x}^c$ is an open set containing $0$, hence ${x} in F$.






    share|cite|improve this answer














    It contains ${0}$ because that is the intersection of the sets $A_n = (-frac{1}{n},frac{1}{n})$, which all verify $0in A_n^circ$, hence $A_n in F$. On the other hand, if $xinmathbb{R}setminus {0}$, then we have that ${x}$ is closed and ${x}^c$ is an open set containing $0$, hence ${x} in F$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 26 at 4:06

























    answered Nov 25 at 19:51









    mlerma54

    1,087138




    1,087138








    • 2




      This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
      – Andreas Blass
      Nov 26 at 1:35












    • You are right, I have edited my solution to make it simpler according to your remark.
      – mlerma54
      Nov 26 at 3:30














    • 2




      This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
      – Andreas Blass
      Nov 26 at 1:35












    • You are right, I have edited my solution to make it simpler according to your remark.
      – mlerma54
      Nov 26 at 3:30








    2




    2




    This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
    – Andreas Blass
    Nov 26 at 1:35






    This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
    – Andreas Blass
    Nov 26 at 1:35














    You are right, I have edited my solution to make it simpler according to your remark.
    – mlerma54
    Nov 26 at 3:30




    You are right, I have edited my solution to make it simpler according to your remark.
    – mlerma54
    Nov 26 at 3:30











    1














    Every $Ssubset Bbb R$ belongs to $sigma(F).$



    (I)... $Ssetminus (-1/n,1/n)in F$ for each $nin Bbb N$ so $Ssetminus {0}=cup_{nin Bbb N},(,Ssetminus (-1/n,1/n),)in sigma (F).$



    (II)... $(-1/n,1/n)in F$ for each $nin Bbb N$ so ${0}=cap_{nin Bbb N}(-1/n,1/n)in sigma (F).$



    (III)... So either $S=Ssetminus {0}in sigma(F)$ or $S=(Ssetminus {0})cup {0}in sigma(F).$






    share|cite|improve this answer


























      1














      Every $Ssubset Bbb R$ belongs to $sigma(F).$



      (I)... $Ssetminus (-1/n,1/n)in F$ for each $nin Bbb N$ so $Ssetminus {0}=cup_{nin Bbb N},(,Ssetminus (-1/n,1/n),)in sigma (F).$



      (II)... $(-1/n,1/n)in F$ for each $nin Bbb N$ so ${0}=cap_{nin Bbb N}(-1/n,1/n)in sigma (F).$



      (III)... So either $S=Ssetminus {0}in sigma(F)$ or $S=(Ssetminus {0})cup {0}in sigma(F).$






      share|cite|improve this answer
























        1












        1








        1






        Every $Ssubset Bbb R$ belongs to $sigma(F).$



        (I)... $Ssetminus (-1/n,1/n)in F$ for each $nin Bbb N$ so $Ssetminus {0}=cup_{nin Bbb N},(,Ssetminus (-1/n,1/n),)in sigma (F).$



        (II)... $(-1/n,1/n)in F$ for each $nin Bbb N$ so ${0}=cap_{nin Bbb N}(-1/n,1/n)in sigma (F).$



        (III)... So either $S=Ssetminus {0}in sigma(F)$ or $S=(Ssetminus {0})cup {0}in sigma(F).$






        share|cite|improve this answer












        Every $Ssubset Bbb R$ belongs to $sigma(F).$



        (I)... $Ssetminus (-1/n,1/n)in F$ for each $nin Bbb N$ so $Ssetminus {0}=cup_{nin Bbb N},(,Ssetminus (-1/n,1/n),)in sigma (F).$



        (II)... $(-1/n,1/n)in F$ for each $nin Bbb N$ so ${0}=cap_{nin Bbb N}(-1/n,1/n)in sigma (F).$



        (III)... So either $S=Ssetminus {0}in sigma(F)$ or $S=(Ssetminus {0})cup {0}in sigma(F).$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 at 21:05









        DanielWainfleet

        34k31647




        34k31647






























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